$(\frac{x}{a})^n + (\frac{y}{b})^n = 1$ as $n \to \infty$

Question

To analyse what happens to the function $$\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 1$$ when as $n \to \infty$.

Here we are given, $a=4,b=10$. When $n=1,2$ I know it is a straight line and ellispe. What happens when $n \to \infty$?

I have plotted it and saw that it converges to rectangle, but can't explain it mathematically!

Answer 1

I think intuitively you could think like this.

There is an assumption that $-a\le x\le a$ and $-b \le y \le b$.

Transform to new coordinates $x'=x/a$ and $y'=y/b$ leading to $$(x')^n + (y')^n = 1$$ $$-1 \le x' \le 1, -1 \le y' \le 1$$ For all $|x'| \lt1$ $$ \lim x'\to 0, n \to\infty $$ Similarly $y'$.

The only non-zero limits are for $x'=\pm1$ and $y'=\pm1$ i.e. when $x=\pm a$ and $y=\pm b$

Answer 2

Yes, it converges to the rectangle in the following sense:

Let $p = (x, y)$ be a point in the curve $\left( \frac xa \right)^n + \left(\frac yb\right)^n = 1$.

Then either $\left( \frac xa\right)^n\ge \frac 12$ or $\left( \frac yb\right)^n \ge \frac 12$, which is

$$ \left|x\right| \ge \sqrt[n]{\frac 12}|a| \ \ \text{or } \left|y\right| \ge \sqrt[n]{\frac 12}|b|. $$

Thus the distance between $p$ and the rectangle $\{ |x| = |a|, |y|= |b|\}$ is less then

$$ \left( 1- \sqrt[n]{\frac 12}\right) \min\{|a|, |b|\}\to 0$$

as $n\to \infty$. In the above sense the curve converges to the rectangle as $n\to \infty$.

One rigorous way is to use the Hausdorff distance between the curve and rectangle (as subsets in $\mathbb R^2$) and show that the Hausdorff disatnce tends to zero.